Lets notice that by Zsigmondy if the exponent of n, in this case is 6 then n can only be 2 such that n^6 -1 doesn't have a prime that doesn't divide n^2 -1 nor n^3 -1. So n=2 only answer. (AOPS still won't let me post LaTeX code notation)

So all prime divisors of $\frac{n^6-1}{(n^3-1)(n+1)}=n^2-n+1$ divide $(n^2-1)(n^3-1)$ Let $p|n^2-n+1$ If $p|n^3-1 \to p| (n+1)(n^2-n+1)-(n^3-1)=2 \to p=2$ but $2\not |n^2-n+1$ So $p|n^2-1 \to p|n-2 \to p|3 \to p=3$ So $n^2-n+1=3^k$ But checking $\pmod {9}$ we can see that $9\not |n^2-n+1$ for any $n$ and so $n^2-n+1=3 \to n=2$